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EXERCISE 2.3 · Q56

Q.If the first term of the G.P. is 6 and its sum to infinity is 9617\dfrac{96}{17}, find the common ratio.

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61−r=9617⇒1−r=6×1796=10296=1716⇒r=1−1716=−116\dfrac6{1-r}=\dfrac{96}{17} \Rightarrow 1-r=\dfrac{6\times17}{96}=\dfrac{102}{96}=\dfrac{17}{16} \Rightarrow r=1-\dfrac{17}{16}=-\dfrac1{16}. Check: 61−(−1/16)=617/16=9617\dfrac6{1-(-1/16)}=\dfrac6{17/16}=\dfrac{96}{17}, confirming the ans …

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